Answer:
Fn = 1,78 [N]
β = 6,4⁰ north-east direction
Step-by-step explanation:
As all charges are positive, forces between electric charges located in the cornes of the base of the triangle and the charge located in the top of the isoceles triangle, are repulsion forces. The east-west direction will depend on where the biggest charge is located, if Q₁ = 9 microC, is in the left end of the triangle base then force on charge Q₃ = 3 microC will have direction north-east, if the above mention configuration change (that means Q₁ is in the right end of the triangle base) then the force will have north-west direction. As in the answer solution, 1,75 [N] direction north-east is offer, we will choose the first configuration. According to that:
K = 1/4*π*ε₀ K = 8,98*10⁹ [N*m²/C²]
The angles in the base of the triangle are equals (we call it α )
sinα = 0,42/045 = 0,93
α = 69⁰
cosα = 0,356
And Q₃ = 3 mC = 3*10⁻⁶ C
Q₁ = 9 mC = 9*10⁻⁶ C
Q₂ = 5 mC = 5*10⁻⁶ C
F₁ [Q₁,Q₃] = K*Q₁*Q₃/d² ⇒F [Q₁,Q₃] = 8,98*10⁻⁹ * 9*10⁻⁶ * 3*10⁻⁶/(0,45)²[N]
F₁ [Q₁,Q₃] = 1,2 [N] direction along the side of the triangle therefore
its components are:
F₁ₓ = F₁* cosα ⇒ F₁ₓ = 1,2 * 0,356
F₁ₓ = 0,43 [N]
F₁y = F₁ *sinα ⇒ F₁y = 1,2 * 0,93
F₁y = 1,16 [N]
Now
F₂ [Q₂,Q₃] = K*Q₂*Q₃/d² ⇒F [Q₂,Q₃] = 8,98*10⁻⁹ * 5*10⁻⁶ * 3*10⁻⁶/(0,45)²[N]
F₂ = 0,67 [N]
F₂ₓ = F₂* cosα ⇒ F₂ₓ = 0,67*0,356
F₂ₓ = 0,24 [N]
F₂y = F₂*sinα ⇒ F₂y = 0,67*0,93
F₂y = 0,62 [N]
We must remember F₁ and F₂ are repulsion forces, then components over y-axis have the same sense (we must add then) and components over x-axis have to be subtracted, in this case, the sense is determine by major force component (to the right) according to this:
Fy = F₁y * F₂y = 1,16 + 0,62
Fy = 1,78 [N]
And
Fₓ = F₁ₓ - F₂ₓ = 0,43 - 0,24
Fₓ = 0,19
Net force module is:
Fn = √ (1,78)² + (0,19)²
Fn = 1,79 [N]
Having an angle respect the y-axis:
tan β = 0,19/1,78 ⇒ tan β = 0,11 and from tan tables
β = 6,4⁰ north-west direction